kushal.adhia wrote:Is q > t ?

(1) qp^2 < tp^2

(2) qp^3 > tp^3

I know this is a simple question but i would really appreciate some help on this...

This is the way i worked it out:

Stmt 1: divide both sides by P^2;
Q>T
Answer is a definite no.

Stmt 2: divide both sides by p^3;
Q>T
Answer is a definite yes.

Hence i chose D... but OA is A

where did i go wrong??

Thanks

Kushal

Source: MGMAT

kmittal82 wrote:Careful, don't just cancel out terms on both sides.

1) qp^2 - tp^2 < 0

p^2(q-t) < 0

p^2 can never be <0, thus q-t < 0 => q< t

Sufficient

2)

qp^3 - tp^3 < 0
p^3(q-t) < 0

If p^3 is < 0, then q > t
If p^3 is > 0, then q < t

Not sufficient

vijchid wrote:Just to show why you cannot cancel the terms

2*-8> 3*-8

if you cancel out the terms -8 you will get 2>3...hope this helps...So when you are not sure about the sign of the number, you cannot cancel out. If you cancel out a negative number, you would have to reverse the inequality.